I am following this paper by Krall and Zettl. I am trying to use the results of Sturm-Louiville (SL) theory to study eigen functions of the classical Legendre equation:

$$ \tag1 \frac{d}{dx}\left((1-x^2)\frac{du}{dx}\right)=\lambda u $$

This SL problem is singular at the points $x=\pm1$ as around these points $(1-x^2)^{-1}$ is not locally integrable. I am trying to show that the end points are limit-circle. i.e. that for all solutions $u$, of $(1)$ that

$$\tag 2u\in L^2(-1,\beta) \quad \forall \beta \in (-1,1) \quad\text{ ( so $x=-1$ is limit circle)}$$

$$ \tag 3u\in L^2(\alpha,1) \quad \forall \alpha \in (-1,1) \quad\text{ ( so $x=1$ is limit circle)}$$

I know the solutions of $(1)$ are given by the Legendre functions $P_\lambda(x),Q_\lambda(x)$ and so I think showing $(2)$ and $(3)$ is equivalent to showing that the $L^2$ norms for both (??) $P_\lambda(x)$ and $Q_\lambda(x)$ are finite, i.e. showing

$$\int^\beta_{-1} P_\lambda(x)^2 dx, \int^\beta_{-1} Q_\lambda(x)^2 dx \quad\text{and}\quad\int^1_{\alpha} P_\lambda(x)^2 dx,\int^1_{\alpha} Q_\lambda(x)^2 dx$$ are finite.

Here is where I get confused, it is stated in the above linked paper that the SL problem is indeed limit circle at both endpoints.


  • If $\lambda$ is taken to be an integer then $P_\lambda(x)$ is a finite degree polynomial and $||P_\lambda||<\infty$ is obvious. However $Q_\lambda(x)$ can still be singular in this case... Is it the case that $Q_\lambda(x)$ diverges slow enough as $x \to \pm1$ to have a finite $L^2$ norm?
  • In the case $\lambda$ is not an integer (this case I am most interested in), the behaviour of $P_\lambda$ and $Q_\lambda$ only get more singular around $x=\pm1$.

In the linked article by Krall et.al. it is mentioned that the classification of boundary points into limit circle / limit point is independent of $\lambda$ so I am assuming there must be some argument in the non-integer $\lambda$ case which shows the poles of $P_\lambda$ and $Q_\lambda$ diverge sufficiently slowly to be in $L^2$.

My first question is how can I show this? I suspect it might be possible by considering the representations of $P_\lambda$ and $Q_\lambda$ in terms of the hypergeometric series, however I have little experience with this and am worried about $_2F_1$ diverging when $x=-1$ since I think this is outside of $_2F_1$'s radius of convergence... Maybe there is an asymptotic argument?

My second question concerns the study of the associated Legendre equation, where the solutions are given by the associated Legendre functions $P^\mu_\lambda(x)$,$Q^\mu_\lambda(x)$.

In this case are the boundary points still limit circle? To show it I think there might be a different weight function $w(x)$ but I think the procedure should be more or less then same.

Would the spectrum still be discrete for the non-integer associated Legendre equation? (for example when the boundary conditions do not require bounded solution at the end points but only $L^2$)

Thank-you for any insight or help you are able to provide!


1 Answer 1


In order to show that the Legendre equation is in the limit circle case, take a look at the solutions of $$ \frac{d}{dx}\left((1-x^2)\frac{df}{dx}\right)=0. $$ This has explicit, classical solutions involving two constants, $A$, $B$: $$ (1-x^2)\frac{df}{dx} = A \\ \frac{df}{dx} = A\frac{1}{1-x^2}=\frac{A}{2}\left[\frac{1}{1-x}+\frac{1}{1+x}\right] \\ f = A\left[-\frac{1}{2}\ln(1-x)+\frac{1}{2}\ln(1+x)\right]+B $$ All solutions are square integrable on $(-1,1)$, regardless of the choice of $A$ and $B$. So the Legendre equation is in the limit-circle case at both $x=-1$ and $x=1$. That means that the classical Legendre equation requires an endpoint condition at $x=-1$ and at $x=1$. It turns out that requiring boundedness near $x=\pm 1$ gives a Sturm-Liouville problem where the only eigenfunctions are the classical Legendre polynomials.

Reference: Selfadjoint Restrictions of Legendre Operator $-\frac{d}{dx}(1-x^{2})\frac{d}{dx}$

  • $\begingroup$ Oh yes, since the LC classification is independent of $\lambda$ we should just take it equal to zero to get the much simpler equation in your answer and solve that. Thanks for your answer! $\endgroup$ Nov 21, 2021 at 9:29
  • $\begingroup$ @valcofadden : Yes, that is correct. And you can find a basis of endpoint functionals that are continuous on the graph of $L$ by using the null space vectors $f=A\left[\cdots\right]+B$ shown above. These endpoint functionals determine all possible self-adjoint versions of the Legendre operator $Lg=-\frac{d}{dx}\left((1-x^2)\frac{dg}{dx}\right)$ in terms of endpoint conditions on $g$ at $\pm 1$. $\endgroup$ Nov 22, 2021 at 5:29

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